POJ3686 The Windy’s
Description
The Windy’s is a world famous toy factory that owns M top-class workshop to make toys. This year the manager receives N orders for toys. The manager knows that every order will take different amount of hours in different workshops. More precisely, the i-th order will take Zij hours if the toys are making in the j-th workshop. Moreover, each order’s work must be wholly completed in the same workshop. And a workshop can not switch to another order until it has finished the previous one. The switch does not cost any time.
The manager wants to minimize the average of the finishing time of the N orders. Can you help him?
Input
The first line of input is the number of test case. The first line of each test case contains two integers, N and M (1 ≤ N,M ≤ 50).
The next N lines each contain M integers, describing the matrix Zij (1 ≤ Zij ≤ 100,000) There is a blank line before each test case.
Output
For each test case output the answer on a single line. The result should be rounded to six decimal places.
Sample Input
3
3 4
100 100 100 1
99 99 99 1
98 98 98 1
3 4
1 100 100 100
99 1 99 99
98 98 1 98
3 4
1 100 100 100
1 99 99 99
98 1 98 98
Sample Output
2.000000
1.000000
1.333333
#include<iostream>
#include<vector>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
using namespace std;
#define INF 0x3f3f3f3f
#define N 10010
struct Edge{
int u,v,cap,flow,cost;
Edge(int xu,int xv,int xcap,int xflow,int xcost){
u=xu;v=xv;cap=xcap;flow=xflow;cost=xcost;
}
};
struct MCMF{
int s,t;
int d[N],f[N],p[N];
bool inq[N];
vector<Edge> E;
vector<int> G[N];
void clear(){
E.clear();
for(int i=0;i<N;i++)G[i].clear();
}
void add(int u,int v,int cap,int cost){
Edge w1(u,v,cap,0,cost);
Edge w2(v,u,0,0,-cost);
E.push_back(w1);
E.push_back(w2);
int m=E.size();
G[u].push_back(m-2);
G[v].push_back(m-1);
}
bool SPFA(int &flow,int &cost){
memset(inq,0,sizeof(inq));
memset(d,0x3f,sizeof(d));
queue<int> Q;
Q.push(s);
d[s]=0;f[s]=INF;
while(!Q.empty()){
int u=Q.front();Q.pop();inq[u]=0;
for(int i=0;i<G[u].size();i++){
Edge e=E[G[u][i]];
if(e.cap>e.flow&&d[e.v]>d[u]+e.cost){
d[e.v]=d[u]+e.cost;
p[e.v]=G[u][i];
f[e.v]=min(f[u],e.cap-e.flow);
if(!inq[e.v]){
Q.push(e.v);
inq[e.v]=1;
}
}
}
}
if(d[t]==INF)return false;
flow+=f[t];cost+=f[t]*d[t];
int u=t;
while(u!=s){
E[p[u]].flow+=f[t];
E[p[u]^1].flow-=f[t];
u=E[p[u]].u;
}
return true;
}
int Min_cost_Max_flow(){
int flow=0,cost=0;
while(SPFA(flow,cost));
return cost;
}
}mcmf;
int n,m,a[100];
int main(){
int T;scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&m);
mcmf.clear();
mcmf.s=0;
mcmf.t=n+n*m+1;
for(int i=1;i<=n;i++){
mcmf.add(0,i,1,0);
for(int j=1;j<=m;j++){
int p;scanf("%d",&p);
for(int k=1;k<=n;k++)
mcmf.add(i,j*n+k,1,k*p);
}
}
for(int i=n+1;i<=n*m+n;i++)mcmf.add(i,n*m+n+1,1,0);
printf("%.6lf
",mcmf.Min_cost_Max_flow()*1.0/n);
}
return 0;
}